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JEE Advanced 2023 Paper 1 · Mathematics · Matrices and Determinants · medium

Let a,b,c be the roots of x^3-3x^2+2x-1=0 and let A=beginpmatrixa&b&c\c&a&b\b&c&aendpmatrix. Then det (adj(A) A^T equals

Single correct option

Let a,b,ca,b,c be the roots of x3−3x2+2x−1=0x^{3}-3x^{2}+2x-1=0 and let A=(abccabbca)A=\begin{pmatrix}a&b&c\\c&a&b\\b&c&a\end{pmatrix}. Then det⁡ ⁣(adj⁡(A) AT)\det\!\big(\operatorname{adj}(A)\,A^{T}\big) equals
  1. A. 729729
  2. B. 8181
  3. C. 65616561
  4. D. 243243
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Answer: (A) 729729

Expanding the circulant determinant gives det⁡A=a3+b3+c3−3abc=(a+b+c)[(a+b+c)2−3(ab+bc+ca)]\det A=a^3+b^3+c^3-3abc=(a+b+c)\big[(a+b+c)^2-3(ab+bc+ca)\big].
By Vieta, a+b+c=3a+b+c=3 and ab+bc+ca=2ab+bc+ca=2, so det⁡A=3(9−6)=9\det A=3(9-6)=9 (this holds for complex roots as well).
For a 3×33\times3 matrix, det⁡(adj⁡A)=(det⁡A)2=81\det(\operatorname{adj}A)=(\det A)^{2}=81 and det⁡AT=det⁡A=9\det A^{T}=\det A=9.
Hence det⁡(adj⁡(A)AT)=81⋅9=729\det(\operatorname{adj}(A)A^{T})=81\cdot9=729.
Note that the determinant formula needs no knowledge of the actual roots, which are one real and two complex numbers. The value 8181 is only det⁡adj⁡A\det\operatorname{adj}A, 6561=946561=9^4 uses the wrong exponent, and 243=81⋅3243=81\cdot3 arises from using e1=3e_1=3 in place of det⁡AT\det A^{T}.
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